栈实现-迷宫问题

给定一个迷宫。如何走出去？最近有看看栈和队列

就想用栈和队列实现，写了半天，终于写出来了。。。

// 0 代表此路可走 1 代表此路不通// 只能走四个方向，即上下左右//周围的 111... 代表围墙。。#include<stdio.h>#include<stdlib.h>#define mac 100//定义栈的大小限制typedef struct{short int row;//行short int col;//列short int dir;//方向}element;element stack[mac];typedef struct{short int vert;//垂直方向short int horiz;//水平方向}offset;offset move[4] = {{-1,0},{0,1},{1,0},{0,-1}};//4个方向的偏移量int top = -1;int maze[10][10] ={{1,1,1,1,1,1,1,1,1,1},{1,0,1,0,0,0,1,1,0,1},{1,0,0,1,1,0,0,1,1,1},{1,1,0,0,0,1,1,0,0,1},{1,0,0,0,0,0,0,0,0,1},{1,1,0,0,0,0,1,1,1,1},{1,1,1,0,0,0,0,0,0,1},{1,1,1,1,0,0,0,0,1,1},{1,1,1,0,0,0,0,0,0,1},{1,1,1,1,1,1,1,1,1,1}};int mark[10][10];#define enter_row 1#define enter_col 1#define exit_row 8#define exit_col 8void path();void push(int *top,element a){++(*top);if(*top <= mac - 1)stack[*top] = a;else{printf("over flow!!!\n");exit(1);}}element pop(int *top){if(*top >= 0)return (stack[(*top)--]);else{printf("stack overflow!!!\n");exit(1);}}element gettop(int top){if(top>=0)return (stack[top]);else{printf("stack undefined.\n");exit(1);}}void path(){int i,j;int row,col,next_row,next_col;int dir,found = 0;//found 标记参数 0代表未访问 1代表访问element position;for(i=0;i<10;i++)for(j=0;j<10;j++)mark[i][j] = 0;mark[1][1] = 1;//初始化标记地图top = 0;stack[0].row = enter_row;//将入口推进栈里面stack[0].col = enter_col;stack[0].dir = 0;while(top>-1&&!found){position = gettop(top);//获取栈顶元素row = position.row;col = position.col;dir = position.dir;while(dir<4&&!found){//没有发现出口next_row = row + move[dir].vert;next_col = col + move[dir].horiz;//探寻下一个方向，首先是上面if(next_row == exit_row && next_col == exit_col)found = 1;//找到出口//如果找到下一个可以走的地方将下一个地方作为新的入口else if(!maze[next_row][next_col] && !mark[next_row][next_col]){mark[next_row][next_col] = 1;position.row = next_row;  position.col = next_col;position.dir = ++dir;push(&top,position);row = next_row; col = next_col; dir = 0;}//找不到，重新试探方向else ++dir;}//再找不到说明死路，退栈if(!found)position = pop(&top);}//找到的话就输出寻找的痕迹 1 为探索过if(found == 1){mark[8][8]=1;printf("探索过的地方为:\n");for(i=0;i<10;i++){for(j=0;j<10;j++)printf("%-2d ",mark[i][j]);printf("\n");}}if(found == 1){printf("     行走的路径为：\n");printf("  步数   行号  列号.\n");for(i=0;i<top;i++)printf("%5d %5d %5d\n",i+1,stack[i].row,stack[i].col);printf("%5d %5d %5d\n",i+1,exit_row,exit_col);}elseprintf("the maze don't have a path.\n");}int main(){path();return 0;}